# SUM

• Apr 28th 2013, 01:20 PM
abualabed
SUM
• Apr 28th 2013, 01:50 PM
$\sum_{n=1}^{20}n(21-n)$
You can use the special sum $\sum\limits_{k = 1}^n {{k^2}} = \frac{{n(n + 1(2n + 1)}}{6}$.
Write your sum as $\sum\limits_{k = 1}^{20} {k(21 - k)} = 21\sum\limits_{k = 1}^{20} k - \sum\limits_{k = 1}^{20} {{k^2}}~.$